wilson-gauss-trifactorial.sf

#!/usr/bin/ruby

#~ func gauss(m) {
    #~ prod(1..m, {|k|
        #~ if (gcd(k, m) == 1) {
            #~ k
        #~ }
        #~ else {
            #~ 1
        #~ }
    #~ })
#~ }

#~ say gauss(13**2)

#~ var p = 43
#~ var t = mfactorial(3*p + kronecker(p, 3), 3)
#~ var g = gcd(t, gauss(p**2))

#~ say t/g

#~ __END__

#~ for p in (4..1e6) {
    #~ #if ((2*k - 1)!! % k**2 - k**2 == -k) {
    #~ #if (mfactorial(3*p - kronecker(p, 3), 3) % p**2 - p**2 / 2 == -p) {
    #~ if (mfactorial(3*p + kronecker(p, 3), 3)  -> is_congruent(-p, p**2)) {
        #~ say p

        #~ if (!p.is_prime) {
            #~ die "Counter-example: #{p}"
        #~ }
    #~ }
    #~ elsif (p.is_prime) {
        #~ die "Missed a prime: #{p}"
    #~ }
#~ }

# (3*k + kronecker(k, 3)) (-3)^floor(((3*k + kronecker(k, 3)) - 1)/3) (1 - (3*k + kronecker(k, 3))/3)_floor(((3*k + kronecker(k, 3)) - 1)/3)

#~ for k in (primes(2, 43)) {
    #~ say [k,

        #~ (k-1)! % k,
        #~ mfactorial(3*k + kronecker(k, 3), 3) % k**2 / k,

        #~ #mfactorial(3*k - 2*kronecker(k, 3), 3) % k**2 / k,

        #~ var t = (mfactorial(3*(k+1) - 1, 3) % k**2)

        #~ t.is_congruent(k, k**2)

        #~ #mfactorial(3*k + kronecker(k, 3), 3) % k**2 / k,

        #~ #k! % k**2 / k
    #~ ]
#~ }

#~ __END__


for p in (primes(2, 100)) {
#for p in (3..10) {
    #~ say [
     #~ p, mfactorial(3*p - kronecker(p, 3), 3) % p**2 - p**2 / 2 ,
        #~ mfactorial(3*p - kronecker(p, 3), 3)  -> is_congruent(-2*p, p**2)
     #~ ]

    say [
        p, mfactorial(3*p + kronecker(p, 3), 3) % p**2 - p**2,
           mfactorial(3*p + kronecker(p, 3), 3) -> is_congruent(-p, p**2)
     ]

    #~ say [
        #~ p, mfactorial(3*p + kronecker(p, 3), 3) % p**2 / p,
        #~ #mfactorial(3*p + kronecker(p, 3), 3)  -> is_congruent(-p, p**2)
     #~ ]

    #~ say [
        #~ p, mfactorial(3*p - kronecker(p, 5), 1) % p**3 / p
    #~ ]

    #say [p, mfactorial(3*p - kronecker(p, 3), 3)   ]
    #say [p, mfactorial(3*p - 3, 3) % p**2 ]
}

__END__

#For p > 2, (2p-1)!! == -p (mod p^2) if and only if p is prime.

Hi,

The following statement should also be provable:

For p > 3, f_3(3p + (p/3)) == -p (mod p^2) if and only if p is prime.

where (p/3) is the Jacobi symbol and f_3(n) are the triple factorial numbers: f_3(n) = A007661(n).




#~ By Wilson's theorem, ((p - 1)/2)!^2 == (-1)^((p + 1)/2) (mod p) for each prime number p. Hence, if p == 3 (mod 4), then ((p - 1)/2)! == +-1 (mod p).